So like if I were asking this question to the Mathematics Stack Exchange I would say something like “I know that $operatorname{SO}(3, 1)$ and $operatorname{SO}(4)$ have $operatorname{SO}(3)$ as a subgroup, and probably bigger ones like $operatorname{SL}(4, mathbb R),$ are those the only 4x4 options? Or are there others too?” Here $operatorname{SO}(3)$ is the group of rotations preserving the Euclidean norm, and we're asking what 4x4 matrix groups have that as a subgroup.

In terms of the physics, we attach less importance to what is possible and more importance to what is actual. So these groups are indeed studied by several of my peers but generally in the contexts of quantum field theory where they describe something about the sorts of particles which exist. I’m from a condensed-matter background so I have less ability to answer those things than a particle physicist.

The relativity of simultaneity

So if I reduce the Lorentz transform to its simplest presentation, the idea is that we want to make sure that people agree on the speed that light is going. You imagine that Alice is moving past Bob at some speed $v ll c$, and maybe Alice presses a button which turns on the lights—we call this an event. Now the light that communicates that this has happened, expands outward from Alice as a thin sphere. Let’s say that it’s a sphere for Bob.

In Bob’s understanding, we would say that Alice is off-center from the sphere, the sphere is centered on some origin $0$ and located off at $x^2 + y^2 + z^2 = c^2 t^2$ whereas Alice is off-center at $(x, y, z) = (v t, 0, 0).$ So one edge is closer to Alice at a distance $t/(c + v)$ and the other is further at a distance $t/(c - v).$ But, if we want to make the speed of light constant for everyone, then when we switch to Alice’s frame of reference we need to put Alice at the center of her sphere. All of the other directions $y,z$ they both agree on, the sphere is tangent to Alice's motion: it is purely this $x$-axis where something needs to change.

The core claim of special relativity is a phenomenon called relativity of simultaneity. Everything else will be derived as a second-order consequence of this claim and we can ignore length contraction and time dilation to first order. The claim is that Alice disagrees with Bob about what is simultaneous. Alice agrees that the light passed this point at distance $t/(c + v)$—but thinks that happened about $t(1 - v/c)$ ago. Like, Bob's clock there said $t$ at the time, but it has always been out-of-sync with the clock Alice carries by this amount. And Alice agrees that the light will pass this point at distance $t/(c-v)$, but thinks that will happen at $t(1 + v/c)$ or so from now.

In other words when we are setting both their clocks to the same zero and calling some instant a time $t$ later as “right now,” to first order Bob sees that “at time 0” some clock at $x = ct$ showed time 0 and sees that “right now” that clock is showing time $t$, and the light is just hitting that clock right now. But Alice thinks that “at time zero” it showed a time $vx/c^2$ and “right now” it is showing a time $t + vx/c^2.$ She agrees that the light passed it when it showed time $t$ but she disputes that this time is “right now,” because the clock was not properly synchronized for her to start with. Instead she thinks that this event happened at time $t' = t - vx/c^2$ for her. In relativity two people who are at the same point agree on what “right now” means at that point. But they disagree on what time “right now” means at far-off locations. If I am on Voyager 2, traveling at about 3.3 AU/year towards a distant star 500 light-years away, and you think based on your complicated equations of stellar evolution that this star is going supernova today, then I think that it went supernova nine and a half days ago. Of course neither of us will get to see the result until about 500 years from now but we will both end up being right when we see our respective results, it is just that what “right now” meant to us at that distance was fundamentally different.

Deriving the Lorentz transform

So if you imagine that Bob built a line of clocks that he thought were all in-sync and all showed $0$ at time $t=0$, Alice thinks that the clock at coordinate $x$ is behind where it should be by a factor $x v/c^2$ (or ahead of where it should be, if $x$ is negative and therefore the factor is negative). This also needs to be understood as a fundamental property of acceleration which we did not appreciate before now because the speed of light is so fast. It is just a property of our universe that if you accelerate with acceleration $alpha$ in the $x$-direction you see an effect that is not explainable as a Doppler shift or anything else, where clocks ahead of you by a coordinate $x$ appear to tick faster a rate of $(1 + alpha x/c^2)$ seconds per second (or tick slower if $x$ is negative, you get the picture). Indeed there must be a surface at $x = -c^2/alpha$ where clocks appear to stand still, this is what we call an “event horizon,” signals of light from before a certain distance cannot reach a constantly-accelerating observer in relativity.

In other words before relativity we connected Alice to Bob with the Galilean transformation, which I will write here with $w = ct$ and $beta = v/c$ as $$begin{bmatrix}w'x'y'z'end{bmatrix} = begin{bmatrix}1&0&0&0-beta&1&0&0&0&1&0&0&0&1end{bmatrix} begin{bmatrix}wxyzend{bmatrix},$$ but now we have a theory that to first-order must be instead $$begin{bmatrix}w'x'y'z'end{bmatrix} = begin{bmatrix}1&-beta&0&0-beta&1&0&0&0&1&0&0&0&1end{bmatrix} begin{bmatrix}wxyzend{bmatrix}.$$ Call this matrix $bar L(beta).$ There are now three ways to proceed. One is to construct $bar L(beta) bar L(-beta)$ on the idea that if Bob sees Alice moving forward at velocity $v_x = +beta c,$ then Alice should see Bob moving backwards with velocity $v_x = -beta c$ and so transforming forwards and then backwards should take us back to where we started, so you get the identity matrix $operatorname{diag}(1, 1, 1, 1)$. But it doesn’t here, it takes us to $operatorname{diag}(1-beta^2, 1-beta^2, 1, 1).$ And the idea is to say “well this was a first-order theory, I can fudge the matrix by dividing its first two rows by $1/sqrt{1 - beta^2}$ and that will propagate entirely through the argument to give me $operatorname{diag}(1, 1, 1, 1)$.” And this works, but maybe it’s not the most stable foundation possible. Another approach is to consider light beams that travel in various “train experiments” and work out these factors much more directly as coming from the Pythagorean theorem, $ct$ being a hypotenuse of a right triangle with base $v t$ and some fixed height $h$.

But my favorite is to force the first-order theory to give you the answer. We try to accelerate by some parameter $phi$, in $N$ steps of size $phi/N$, and we therefore form $$L(phi) = lim_{Ntoinfty} [bar L(phi/N)]^N.$$Matrix exponentiation requires an eigenbasis, but an eigenbasis is very easy to come by: $[1, 1, 0, 0]$ and $[-1, 1, 0, 0]$ are clear eigenvectors to join $[0, 0, 1, 0]$ and $[0, 0, 0, 1].$ So one can work out that in fact,$$ L(phi) = begin{bmatrix}coshphi&-sinhphi&0&0-sinhphi&coshphi&0&0&0&1&0&0&0&1end{bmatrix},$$and now $L(phi)L(-phi) = I$ via the rules for hyperbolic sines and hyperbolic cosines that $cosh^2phi - sinh^2phi = 1.$ This connects to the other two approaches by identifying that actually $beta = tanhphi = sinh phi/coshphi,$ at which point one can work out that $cosh^{-2}phi = 1 - tanh^2phi$ and therefore $coshphi = 1/sqrt{1 - beta^2}.$ So the same parameter reappears but in a much more logically rigorous way that assures us that indeed, every other effect in relativity comes from the relativity of simultaneity compounded with the Galilean transformation.

To get the full Lorentz group one composes this boost operation with the rotations and one thereby gets a group of all linear transforms that preserve the Lorentz-norm $w^2 - x^2 - y^2 - z^2.$

I think that’s pretty elegant but the fundamental question of “which other groups of these 4x4 matrices have SO(3) as a subgroup?” I think is also a valid question which you might ask a mathematician.